Question

find the sum of the AP：1+2+3+4..........+99+100​

Answer 1

Answer:

Step-by-step explanation:

=1+2−3+4+…−99+100  

Split this series into 3 as:

S=[1+4+7+...+100]S1+[2+5+8+...+98]S2−[3+6+9+...+99]S3

S1=1+4+7+...+100

This is an AP with A1=1,L1=100,D1=3

Number of terms =N1=L1−A1D1+1=100−13+1=34

S1=N12.[A1+L1]=342.[1+100]=1717

S2=2+5+8+...+98

This is an AP with A2=2,L2=98,D2=3

Number of terms =N2=L2−A2D2+1=98−23+1=33

S2=N22.[A2+L2]=332.[2+98]=1650

S3=3+6+9+...+99

This is an AP with A3=3,L3=99,D3=3

Number of terms =N3=L3−A3D3+1=99−33+1=33

S3=N32.[A3+L3]=332.[3+99]=1683

S=S1+S2−S3=1717+1650−1683=1684

Ans: 1684

Alternate method:

Consider the triplets:

T1=1+2−3=0

T2=4+5−6=3

T3=7+8−9=6

…

T33=97+98−99=96

S=T1+T2+T3+…+T33+100=0+3+6+…+96AP with A= 0, D=3, N=33+100

=332.[0+96]+100=1584+100=1684

hope this helps u

Answer 2

The sum of ap is as follows